Let $R$ be the region enclosed by the polar curve $r(\theta)=4\cos(2\theta)$ where $-\dfrac{\pi}{8}\leq \theta\leq \dfrac{\pi}{8}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize-\dfrac{\pi}{8}}^{\scriptsize\dfrac{\pi}{8}}8\cos^2(2\theta)\,d\theta$ (Choice B) B $ \int_{\scriptsize-\dfrac{\pi}{8}}^{\scriptsize\dfrac{\pi}{8}}4\cos^2(2\theta)\,d\theta$ (Choice C) C $ \int_0^{\scriptsize-\dfrac{\pi}{4}}8\cos^2(2\theta)\,d\theta$ (Choice D) D $ \int_0^{\scriptsize-\dfrac{\pi}{4}}4\cos^2(2\theta)\,d\theta$
This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=4\cos(2\theta)}$, ${\alpha=-\dfrac{\pi}{8}}$, and ${\beta=\dfrac{\pi}{8}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize-\dfrac{\pi}{8}}}^{{\scriptsize\dfrac{\pi}{8}}}\dfrac{1}{2}\left({4\cos(2\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize-\dfrac{\pi}{8}}^{\scriptsize\dfrac{\pi}{8}}\dfrac{1}{2}\cdot16\cos^2(2\theta)d\theta \\\\ &= \int_{\scriptsize-\dfrac{\pi}{8}}^{\scriptsize\dfrac{\pi}{8}}8\cos^2(2\theta)\,d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize-\dfrac{\pi}{8}}^{\scriptsize\dfrac{\pi}{8}}8\cos^2(2\theta)\,d\theta$